比例性质

在一个比例等式中，第一个比例的前后项之和与第一个比例的后项的比，等于第二个比例的前后项之和与第二个比例的后项的比。

已知${\displaystyle a,b,c,d\in \mathbb {C} }$，且有${\displaystyle b\neq 0,d\neq 0}$，如果${\displaystyle {\frac {a}{b}}={\frac {c}{d}}}$，则有${\displaystyle {\frac {a+b}{b}}={\frac {c+d}{d}}}$。

${\displaystyle \because {\frac {a}{b}}={\frac {c}{d}}}$

${\displaystyle \therefore {\frac {a}{b}}+1={\frac {c}{d}}+1}$

${\displaystyle \therefore {\frac {a+b}{b}}={\frac {c+d}{d}}}$

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在一个比例等式中，第一个比例的前后项之差与第一个比例的后项的比，等于第二个比例的前后项之差与第二个比例的后项的比。

已知${\displaystyle a,b,c,d\in \mathbb {C} }$，且有${\displaystyle b\neq 0,d\neq 0}$，如果${\displaystyle {\frac {a}{b}}={\frac {c}{d}}}$，则有${\displaystyle {\frac {a-b}{b}}={\frac {c-d}{d}}}$。

${\displaystyle \because {\frac {a}{b}}={\frac {c}{d}}}$

${\displaystyle \therefore {\frac {a}{b}}-1={\frac {c}{d}}-1}$

${\displaystyle \therefore {\frac {a-b}{b}}={\frac {c-d}{d}}}$

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在一个比例等式中，第一个比例的前后项之和与第一个比例的前后项之差的比，等于第二个比例的前后项之和与第二个比例的前后项之差的比。

已知${\displaystyle a,b,c,d\in \mathbb {C} }$，且有${\displaystyle b\neq 0,d\neq 0}$，如果${\displaystyle {\frac {a}{b}}={\frac {c}{d}}}$，则有${\displaystyle {\frac {a+b}{a-b}}={\frac {c+d}{c-d}}}$。

${\displaystyle \because {\frac {a}{b}}={\frac {c}{d}}}$

令${\displaystyle k={\frac {a}{b}}={\frac {c}{d}}}$，则${\displaystyle a=b\cdot k,c=d\cdot k}$

${\displaystyle {\frac {a+b}{a-b}}={\frac {bk+b}{bk-b}}={\frac {k+1}{k-1}}}$

${\displaystyle {\frac {c+d}{c-d}}={\frac {dk+d}{dk-d}}={\frac {k+1}{k-1}}}$

${\displaystyle \therefore {\frac {a+b}{a-b}}={\frac {c+d}{c-d}}}$

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在一个比例等式中，两前项之和与两后项之和的比例与原比例相等

已知${\displaystyle a,b,c,d\in \mathbb {C} }$，且有${\displaystyle b\neq 0,d\neq 0}$，如果${\displaystyle {\frac {a}{b}}={\frac {c}{d}}}$，则有${\displaystyle {\frac {a+c}{b+d}}={\frac {a}{b}}={\frac {c}{d}}}$。

${\displaystyle \because {\frac {a}{b}}={\frac {c}{d}}}$

令${\displaystyle k={\frac {a}{b}}={\frac {c}{d}}}$，则${\displaystyle a=b\cdot k,c=d\cdot k}$

${\displaystyle {\frac {a+c}{b+d}}={\frac {bk+dk}{b+d}}=k}$

${\displaystyle \therefore {\frac {a+c}{b+d}}={\frac {a}{b}}={\frac {c}{d}}}$

证毕

${\displaystyle \because {\frac {a}{b}}={\frac {c}{d}}}$

${\displaystyle \therefore {\frac {a}{c}}={\frac {b}{d}}}$

由合比性质${\displaystyle {\frac {a+c}{c}}={\frac {b+d}{d}}}$

${\displaystyle \therefore {\frac {a+c}{b+d}}={\frac {c}{d}}}$

即${\displaystyle {\frac {a+c}{b+d}}={\frac {a}{b}}={\frac {c}{d}}}$

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已知${\displaystyle a_{i},b_{i}\in \mathbb {C} ,i=1,2,\ldots ,n}$，且有${\displaystyle b_{i}\neq 0}$，如果${\displaystyle {\frac {a_{1}}{b_{1}}}={\frac {a_{2}}{b_{2}}}=\ldots ={\frac {a_{n}}{b_{n}}}}$，则有${\displaystyle {\frac {\sum _{i}^{n}{a_{i}}}{\sum _{i}^{n}{b_{i}}}}={\frac {a_{i}}{b_{i}}}}$。

已知${\displaystyle a,b,c,d\in \mathbb {C} }$，且有${\displaystyle b\neq 0,d\neq 0}$，如果${\displaystyle {\frac {a}{b}}={\frac {c}{d}}}$，则有${\displaystyle {\frac {\lambda a+\mu b}{\xi a+\eta b}}={\frac {\lambda c+\mu d}{\xi c+\eta d}}}$。

证明：

${\displaystyle \because {\frac {a}{b}}={\frac {c}{d}}}$

令${\displaystyle k={\frac {a}{b}}={\frac {c}{d}}}$，则${\displaystyle a=b\cdot k,c=d\cdot k}$

${\displaystyle {\frac {\lambda a+\mu b}{\xi a+\eta b}}={\frac {\lambda bk+\mu b}{\xi bk+\eta b}}={\frac {\lambda k+\mu }{\xi k+\eta }}}$

${\displaystyle {\frac {\lambda c+\mu d}{\xi c+\eta d}}={\frac {\lambda dk+\mu d}{\xi dk+\eta d}}={\frac {\lambda k+\mu }{\xi k+\eta }}}$

${\displaystyle \therefore {\frac {\lambda a+\mu b}{\xi a+\eta b}}={\frac {\lambda c+\mu d}{\xi c+\eta d}}}$

证毕

已知${\displaystyle a,b,c,d\in \mathbb {C} }$，且有${\displaystyle b\neq 0,d\neq 0}$，如果${\displaystyle {\frac {a}{b}}={\frac {c}{d}}}$，则有${\displaystyle {\frac {\lambda a+\mu c}{\lambda b+\mu d}}={\frac {a}{b}}={\frac {c}{d}}}$。

证明：

${\displaystyle \because {\frac {a}{b}}={\frac {c}{d}}}$

令${\displaystyle k={\frac {a}{b}}={\frac {c}{d}}}$，则${\displaystyle a=b\cdot k,c=d\cdot k}$

${\displaystyle {\frac {\lambda a+\mu c}{\lambda b+\mu d}}={\frac {\lambda bk+\mu dk}{\lambda b+\mu d}}=k}$

${\displaystyle \therefore {\frac {\lambda a+\mu c}{\lambda b+\mu d}}={\frac {a}{b}}={\frac {c}{d}}}$

证毕

已知${\displaystyle a_{i},b_{i}\in \mathbb {C} ,i=1,2,\ldots ,n}$，且有${\displaystyle b_{i}\neq 0}$，如果${\displaystyle {\frac {a_{1}}{b_{1}}}={\frac {a_{2}}{b_{2}}}=\ldots ={\frac {a_{n}}{b_{n}}}}$，则有${\displaystyle {\frac {\sum _{i}^{n}{\lambda _{i}a_{i}}}{\sum _{i}^{n}{\lambda _{i}b_{i}}}}={\frac {a_{i}}{b_{i}}}}$。

证明：

令${\displaystyle k={\frac {a_{i}}{b_{i}}}}$，则${\displaystyle a_{i}=b_{i}\cdot k}$

${\displaystyle {\frac {\sum _{i}^{n}{\lambda _{i}a_{i}}}{\sum _{i}^{n}{\lambda _{i}b_{i}}}}={\frac {\sum _{i}^{n}{\lambda _{i}kb_{i}}}{\sum _{i}^{n}{\lambda _{i}b_{i}}}}={\frac {k\sum _{i}^{n}{\lambda _{i}b_{i}}}{\sum _{i}^{n}{\lambda _{i}b_{i}}}}=k}$

${\displaystyle \therefore {\frac {\sum _{i}^{n}{\lambda _{i}a_{i}}}{\sum _{i}^{n}{\lambda _{i}b_{i}}}}={\frac {a_{i}}{b_{i}}}}$

证毕